Python 将返回类型注释为赋予函数的类型

Question

I am giving a type to function. The function instantiates the type for me. The type implements BaseCustomObject but is one of many possible types.

def factory(object_to_create:[T], parameter:int=1) -> [T]:
    constructor_parameter = calculation(parameter)
    return object_to_create(constructor_parameter)

my_custom_object_instance = factory(MyCustomObject)

How do I annotate factory so that it is clear that it gives back whatever type I send it, and not just BaseCustomObject ? I prefer not to use a Union type but to have the annotation/inspection/ide understand exactly which specific type I am getting back.

Answer 1

use a TypeVar generic type. [T] means a list of T, not a T, so dont do that.

If object_to_create is a function, do this.

# Use bound to limit the type to a specific ancestor.
T = TypeVar("T", bound="BaseCustomObject") 

# Callable[..., T] is the type of a function 
# that takes in whatever argument and returns an instance of T.

def my_func(
  object_creator: Callable[..., T], 
  params: Optional[int] = 1
) -> T:
  return object_creator(params) # Will return a T.

If object_to_create is a class and not a function you instead use Type[T] which will be the actual class and not a class instance.

# Use bound to limit the type to a specific ancestor.
T = TypeVar("T", bound="BaseCustomObject") 

def my_func(
  object_to_create: Type[T], 
  params: Optional[int] = 1
) -> T:
  return object_to_create(params) # Will return an instance of object_to_create.